Problem: There are N men and N women, both numbered 1, 2, \ldots, N.

For each i, j (1 \leq i, j \leq N), the compatibility of Man i and Woman j is given as an integer a_{i, j}. If a_{i, j} = 1, Man i and Woman j are compatible; if a_{i, j} = 0, they are not.

Taro is trying to make N pairs, each consisting of a man and a woman who are compatible. Here, each man and each woman must belong to exactly one pair.

Find the number of ways in which Taro can make N pairs, modulo 10^9 + 7.

Constraints

* All values in input are integers.
* 1 \leq N \leq 21
* a_{i, j} is 0 or 1.

Input

Input is given from Standard Input in the following format:


N
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}


Output

Print the number of ways in which Taro can make N pairs, modulo 10^9 + 7.

Examples

Input

3
0 1 1
1 0 1
1 1 1


Output

3


Input

4
0 1 0 0
0 0 0 1
1 0 0 0
0 0 1 0


Output

1


Input

1
0


Output

0


Input

21
0 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1
1 1 1 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 1 0
0 0 1 1 1 1 0 1 1 0 0 1 0 0 1 1 0 0 0 1 1
0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 0 0 1 1 0
1 1 0 0 1 0 1 0 0 1 1 1 1 0 0 0 0 0 0 0 0
0 1 1 0 1 1 1 0 1 1 1 0 0 0 1 1 1 1 0 0 1
0 1 0 0 0 1 0 1 0 0 0 1 1 1 0 0 1 1 0 1 0
0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1 1
0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1
0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 0 0 1
0 1 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1 0 1 1 0
0 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1
0 1 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 0 1 1
1 1 1 1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 1
0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1
1 0 1 1 0 1 0 1 0 0 1 0 0 1 1 0 1 0 1 1 0
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 0 0 1
0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1
0 0 0 0 1 1 1 0 1 0 1 1 1 0 1 1 0 0 1 1 0
1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 1 1 1 0 1 0
1 0 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 0 0 0


Output

102515160

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

const int N = 22;
const int mod = 1e9 + 7;

int n, a[N][N];
int f[N][1 << N];

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= n; ++j) scanf("%d", &a[i][j]);
	}
	f[0][0] = 1;
	for(int S = 1; S < (1 << n); ++S) {
		int i = 0;
		for(int j = 0; j < n; ++j) if((S >> j) & 1) ++i;
		for(int j = 0; j < n; ++j) {
			if(((S >> j) & 1) && a[i][j + 1]) (f[i][S] += f[i - 1][S ^ (1 << j)]) %= mod;
		}
	}
	printf("%d\n", f[n][(1 << n) - 1]);
}